| ⇦ | 
| ⇨ |
A Carnot engine, having an efficiency of η=1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is
Options
(a) 100 J
(b) 99 J
(c) 90 J
(d) 1 J
Correct Answer:
90 J
Explanation:
Efficiency of carnot engine n = 1 – (T₂ / T₁)
That is, 1 / 10 = 1 – (T₂ / T₁)
⇒ (T₂ / T₁) = 1 – (1 / 10) = 9 / 10 ⇒ (T₁ / T₂) = 10 / 9
.·. w = Q₂ . [(T₁ / T₂) – 1) ⇒ 10 = Q₂ [(10 / 9) – 1]
⇒ 10 = Q₂ (1 / 9) ⇒ Q₂ = 90 J
So, 90 J heat is absorbed at lower temperature.
Related Questions: - Two wires of equal length and equal diameter and having resistivities ρ₁ and ρ₂
- A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance
- In depletion layer of unbiased P-N junction
- Given A=4i+6j and B=2i+3j. Which of the following is correct?
- The speed of light in media M₁ and M₂ are 1.5×10⁸ ms⁻¹ and 2×10⁸ ms⁻¹ respectively.
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two wires of equal length and equal diameter and having resistivities ρ₁ and ρ₂
- A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance
- In depletion layer of unbiased P-N junction
- Given A=4i+6j and B=2i+3j. Which of the following is correct?
- The speed of light in media M₁ and M₂ are 1.5×10⁸ ms⁻¹ and 2×10⁸ ms⁻¹ respectively.
Topics: Thermodynamics (179)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply